Question

Using properties of proportion, solve for x:
$\frac{ \sqrt{x + 1} \: + \: \sqrt{x - 1} }{ \sqrt{x + 1} \: - \: \sqrt{x - 1} } = \frac{4x - 1}{2}$
​

Answer 1

$<b><u>Answer:-</u></b>$

$<u>Properties of proportion:-</u>$

If $\frac{a}{b} = \frac{c}{d} \implies$

1. $\frac{b}{a} = \frac{d}{c}$ ( By Invertendo )

2. $\frac{a}{c} = \frac{b}{d}$ ( By Alternendo )

3. $\frac{a + b}{b} = \frac{c + d}{d}$ ( By Compendo )

4. $\frac{a - b}{b} = \frac{c - d}{d}$ ( By dividendo )

5. $\frac{a + b}{a - b} = \frac{c + d}{c - d}$ ( By Compendo and dividendo )

6. $\frac{a}{a - b} = \frac{c}{c - d}$ ( By Convertendo )

7. If $\frac{a}{b} = \frac{c}{d} = \frac {e}{f}$, then each ratio :

= $\frac{a + c + e}{b + d + f}$

= $\frac{Sum \:of \:antecedents}{Sum\: of \:consequents}$

$<b><u>Solution:-</u></b>$

$<i><u>To Find </u></i>$ : $\frac{ \sqrt{x + 1} \: + \: \sqrt{x - 1} }{ \sqrt{x + 1} \: - \: \sqrt{x - 1} = \frac{4x - 1}{2}$

Applying Compendo and dividendo,

$\frac{ \sqrt{x + 1} \: + \: \sqrt{x - 1} \: + \: \sqrt{x + 1} \: - \: \sqrt{x - 1 }{ \sqrt{x + 1} \: +\: \sqrt{x + 1} \: - \: \sqrt{x + 1} \: + \: \sqrt{x - 1}$ = $\frac{4x - 1 + 2}{4x - 1 - 2}$

$\frac {2 \sqrt{x + 1}}{2 \sqrt{x - 1}} = \frac{4x + 1}{4x - 3}$

Squaring on both sides,

$\frac {x + 1}{x - 1} = \frac {16x^{2} + 1 + 8x}{16x^{2} + 9 - 24x}$

Again Applying Compendo and dividendo,

$\frac{x + 1 + x - 1}{x + 1 - x + 1} = \frac {16x^{2} + 1 + 8x + 16x^{2} + 9 - 24x}{16x^{2} + 1 + 8x -16x^{2} - 9 + 24x}$

$\frac {2x}{2} = \frac {32^{2} + 10 - 16x}{32x - 8}$

⇒ 16x² - 4x = 16x² + 5 - 8x

⇒ 4x = 5

⇒ x = $\frac {5}{4}$

Answer 2

Answer:

Properties of proportion:-

If \frac{a}{b} = \frac{c}{d} \implies

b

a

=

d

c

⟹

1. \frac{b}{a} = \frac{d}{c}

a

b

=

c

d

( By Invertendo )

2. \frac{a}{c} = \frac{b}{d}

c

a

=

d

b

( By Alternendo )

3. \frac{a + b}{b} = \frac{c + d}{d}

b

a+b

=

d

c+d

( By Compendo )

4. \frac{a - b}{b} = \frac{c - d}{d}

b

a−b

=

d

c−d

( By dividendo )

5. \frac{a + b}{a - b} = \frac{c + d}{c - d}

a−b

a+b

=

c−d

c+d

( By Compendo and dividendo )

6. \frac{a}{a - b} = \frac{c}{c - d}

a−b

a

=

c−d

c

( By Convertendo )

7. If \frac{a}{b} = \frac{c}{d} = \frac {e}{f}

b

a

=

d

c

=

f

e

, then each ratio :

= \frac{a + c + e}{b + d + f}

b+d+f

a+c+e

= \frac{Sum \:of \:antecedents}{Sum\: of \:consequents}

Sumofconsequents

Sumofantecedents

Solution:-

To Find : :\frac{ \sqrt{x + 1} \: + \: \sqrt{x - 1} }{ \sqrt{x + 1} \: - \: \sqrt{x - 1} = \frac{4x - 1}{2}

Applying Compendo and dividendo,

\frac{ \sqrt{x + 1} \: + \: \sqrt{x - 1} \: + \: \sqrt{x + 1} \: - \: \sqrt{x - 1 }{ \sqrt{x + 1} \: +\: \sqrt{x + 1} \: - \: \sqrt{x + 1} \: + \: \sqrt{x - 1} = \frac{4x - 1 + 2}{4x - 1 - 2}

\frac {2 \sqrt{x + 1}}{2 \sqrt{x - 1}} = \frac{4x + 1}{4x - 3}

2

x−1

2

x+1

=

4x−3

4x+1

Squaring on both sides,

\frac {x + 1}{x - 1} = \frac {16x^{2} + 1 + 8x}{16x^{2} + 9 - 24x}

x−1

x+1

=

16x

2

+9−24x

16x

2

+1+8x

Again Applying Compendo and dividendo,

\frac{x + 1 + x - 1}{x + 1 - x + 1} = \frac {16x^{2} + 1 + 8x + 16x^{2} + 9 - 24x}{16x^{2} + 1 + 8x -16x^{2} - 9 + 24x}

x+1−x+1

x+1+x−1

=

16x

2

+1+8x−16x

2

−9+24x

16x

2

+1+8x+16x

2

+9−24x

\frac {2x}{2} = \frac {32^{2} + 10 - 16x}{32x - 8}

2

2x

=

32x−8

32

2

+10−16x

⇒ 16x² - 4x = 16x² + 5 - 8x

⇒ 4x = 5

⇒ x = \frac {5}{4}

4

5